In this post, I want to focus on the Riemman $\zeta$ function, and (one type of) its generalizations: the Dirichlet $L$functions. We’ll prove some nice properties of these things (e.g. their meromorphic continuations, product formulas, and functional equations), and use them to prove some infinitude results involving primes (spoiler: we’ll show e.g. that there are infinitely many primes!).
Riemann
We want to show that, among other things, this function extends to a meromorphic function on the entire complex plane. To begin, we’ll show that it’s at least holomorphic in the halfplane $\Re(s)>1$ by appealing to a standard theorem from complex analysis ^{1}.
Next, we’ll derive the product formula for the $\zeta$ function. Morally, we want to perform the following manipulation (where $p$ always denotes a prime because we’re not savages)$\dots$
Above, $v_p(n)$ is the number of times the prime $p$ divides the number $n$. The middle equality is (formally) justified by the fundamental theorem of arithmetic; every number has a unique factorization into primes and this corresponds to picking some exponent for each prime ^{2}. The above is called an Euler product because Euler was the first person to show this equality, but I think his argument was about as legitimate as what I wrote above, so let’s do one better and actually prove this.
$(\le)$ Fix some positive integer $N\ge1$. Then, by the fundamental theorem of arithmetic, we have $$\prod_{p\le N}\sum_{n=0}^N\frac1{p^{sn}}=\sum_{n\in S}\frac1{n^s}\le\sum_{n=1}^\infty\frac1{n^s}=\zeta(s),$$ where $S=\bracks{n:p\le N\implies v_p(n)\le N\text{ and }p>N\implies v_p(n)=0}$. Taking the limit as $N\to\infty$ gives $$\prod_p\sum_{n\ge1}\frac1{p^{sn}}=\prod_p\parens{\frac1{1p^{s}}}\le\zeta(s),$$ which has the added benefit of showing the Euler product converges.
$(\ge)$ Fix some positive integer $N\ge1$. Then, $$\sum_{n=1}^N\frac1{n^s}\le\prod_{p\le N}\sum_{n=0}^N\frac1{p^{ns}}\le\prod_p\sum_{n=0}^\infty\frac1{p^{ns}}.$$ Taking the limit as $N\to\infty$ gives $$\zeta(s)\le\prod_p\parens{\frac1{1p^{s}}}$$ as desired.
i.e let $p_1=2$, and for $n>1$, let $p_n$ be the smallest prime factor of $p_1p_2\cdots p_{n1}+1$. Then, $p_n$ is prime for all $n$, and $n\neq m\implies p_n\neq p_m$, so this gives an infinite sequence of primes. No contradiction necessary.
While we’re on the subject of primes, we can actually do more than just count them.
We next move on to analytically continuing the $\zeta$ function. I wish I could give some good motiviation for the argument, but sadly, I cannot. The main idea is to relate the $\zeta$ function to the theta function from last time
and then translate the equality $\vartheta(s)=s^{1/2}\vartheta(1/s)$ into a similar functional equation for the $\zeta$ function. How do you think up this approach? I don’t know. ^{3}
Before the proof, recall the Gamma function
which is initally defined for $\Re(s)>0$ but extends to a meromorphic function on $\C$ with simple poles at the nonpositive integers. The idea here is that integration by parts gives $\Gamma(s+1)=s\Gamma(s)$, so
gives a meromorphic extension of $\Gamma$ to the half plane $\Re(s)>m$. We can relate $\zeta(s)$ to $\vartheta(s)$ through $\Gamma(s)$ via the following equality:
This is seen by making the change of variables $u=\frac t{\pi n^2}$ on the LHS. Motivated by this, call $\xi(s)=\pi^{s/2}\Gamma(s/2)\zeta(s)$ the xi function ^{4}.
It may be worth noting that the product formula shows that there are no zeros with $\Re(s)>1$, and combining this with the above functional equation shows that the only zeros with $\Re(s)<0$ are at the negative even integers $s=2k$ with $k\in\Z_{\ge1}$. Thus, any “nontrivial” zero of the Zeta function must lie n the strip $0\le\Re(s)\le1$. This leads me to make the following totally 100% original conjecture:
Dirichlet
Dirichlet studied his $L$series with one specific application in mind: proving that for any coprime $a,n$, there are infinitely many primes $p\equiv a\pmod n$ ^{5}. At its core, the idea behind the proof is to follow in Euler’s footsteps by showing that
Unsurprisingly, we will prove this by exploiting some analytic properties of a functions that are reminiscent of $\zeta(s)$.
This summation is similar to $\zeta(1)$, so, letting $\mbf1_ a:\Z\to\bits$ be the characterstic function of congruence to $a\pmod n$, we may be tempted to consider the function $\sum_{n=1}^\infty\mbf1_a(n)n^{s}$. However, $\mbf1_a$ is not multiplicative, so we’d have a hard time recovering an Euler product for this; since Euler products were useful in proving the infinitue of primes, we’d like to still have one of those. On the bright side, $\mbf1_a$ descends to a homomorphism $\units{(\zmod n)}\to\C$.
Our first goal is to relate these characters to $\mbf1_a$, our real function of interest. First, note that any character $\chi:\units{(\zmod N)}\to\C$ can be extended to a function $\Z\to\C$ via
Furthermore, these extensions are completely multiplicative in the sence that $\chi(nm)=\chi(n)\chi(m)$ for all $n,m\in\Z$. To relate this extension to $\mbf1_a:\Z\to\bits\subset\C$, we’ll take a shallow dive in general character theory ^{6}.
Shallow Dive
Fix a finite abelian group $G$. By a character on $G$, we mean a homomorphism $G\to S^1$.
Combining the two remarks and the proposition above in a simple induction argument gives the following.
We can actually say something stronger in the case of double duals.
Finally, we arrive at our last result of this minisection.
The corollary is just applying the theorem to $\hat G$.
Back to Dirichlet
Ok, back to Dirichlet. Remember that we want to relate $\mbf1_a:\Z\to\C$, the characteristic function of being $\equiv a\pmod N$. to characters of the group $\units{(\zmod N)}$. This will be an application of the last theorem from our dive into character theory. First, to make things notationally easier, define $U(N)=\units{(\zmod N)}$ and $X(N):=\wh{U(N)}$.
Hence, to understand
it should suffice to study the Dirichlet $L$series defined below.
Note that the trivial character $\chi=1$ has $\zeta(s)$ as its $L$series (when $N=1$).
Fix some (primitive) character $\chi\in X(N)$ with $N>1$ (so $\chi$ nontrivial!). It will turn out that these $L$series each extend to meromorphic ^{7} functions on the plane with their own product formulae and functional equations. The proofs of these facts are similar to those in the case of the Zeta function, so we may not always be as careful and trust that arguments could be made more carefully. For example, we observe the following product formula
Before moving on, fix $\eps\in\bits$ such that $\chi(1)=(1)^{\eps}$. We call this $\eps$ the exponent of $\chi$. We use it to define the $\chi$Gamma function
Perhaps unsurprisingly, the next thing we do is perform a substitution ($t\mapsto\pi n^2u/N$) to get
Multiplying by $\chi(n)$ and summing over all $n\ge1$ gives
Motivated by this, we also define a $\chi$analogue of the theta series
In the above definition, since $\chi$ is nontrivial, we have $\chi(0)=0$; hence, there is no $n=0$ term above. This may seem like a small remark, but the $n=0$ term in the case of trivial $\chi$ of conductor $N$ (i.e. in the case of Riemann zeta) was the source of the poles of the completed xi function. At this point, we would like to be able to apply Poisson summation ^{8} to get a functional equation for $\theta(\chi, z)$. To do this, first define
Next, we want to calculate the fourier transform of $f_a(x)=(Nx+a)^{\eps}e^{\pi(Nx+a)^2s/N}$ ($\Re(s)>0$). For conveinence, let $g_0(x)=e^{\pi x^2s/N}$ and $g_1(x)=xe^{\pi x^2s/N}$. Then,
With that over, Poisson summation gives ^{9}
so
The sum over $a$ above is a special kind of sum called a Gauss sum, which are sums of the form
 $\tau(\chi, k)=\conj\chi(k)\tau(\chi, 1)$ (the bar denotes complex conjugation).
 $\abs{\tau(\chi)}=\sqrt N$.
 $\tau(\conj\chi, 1)=\chi(1)\conj{\tau(\chi, 1)}=(1)^{\eps}\conj{\tau(\chi, 1)}$.
Now, let’s continue to simplify our expression for $\theta(\chi, s)$. Using the above (and that $i=i^{1}$) we see
where I just did a bunch of algebra, so it’s very possible I made a mistake along the way, and the true expression should look slightly different ^{10}. However, assuming this is not the case ^{11}, we have a nice functional equation for $\theta(\chi, s)$, so we can use this to give an analytic extension of $L(\chi, s)$ in much the same way as before. If you recall, we had shown previously that
Hence, we will find a functional equation for the $\chi$xi function ^{12}
Now that we’ve gotten this far, let’s return to the question of primes in arithmetic progressions. In order to prove Dirichlet’s theorem, we’ll need to make use of one nontrivial result that I will not prove in this post ^{13}
Now, let
and recall that
Combined together, this gives
Now, note that (say, $s>1$)
where I skipped a few steps because the argument is the same as when we showed $\sum_p\inv p=\infty$. Now, here’s the kicker: taking the limit as $s\to1^+$, we get
where the last equality comes from the fact that $\log L(\chi,1)=\infty$ iff $\chi$ is the trivial character! Thus, we’ve proven the following.
Dedekind
Dedekind’s name doesn’t appear in the title because I wasn’t originally going to talk about him, but he has a role in this story too. Unlike the previous two sections, this one will require some knowledge of basic algebraic number theory, and will not prove an infinitude result about primes ^{14}.
We start by recalling some definitions/facts. A number field is a finite extension $K/\Q$. The integral closure of $\Z$ in $K$ is denoted $\ints K$, and is called $K$’s ring of integers. $\ints K$ is always a Dedekind domain (but not always a UFD), so any nonzero ideal in $\ints K$ factors into a unique product of prime ideals. Given an ideal $I\subseteq\ints K$, its norm is $N(I)=\abs{\ints K/I}$, the size of its residue ring.
Now, as it turns out, Dedekind’s favorite letter is the same as Riemann’s.
One can use $\zeta_K$ to prove my earlier claim about the nonvanishing of $L$functions attached to nontrivial characters at $s=1$ ^{15}. This follows as a corollary of the following.
The LHS above has a simple pole at $s=1$. The factor on the RHS belonging to the trivial character (which is just $\zeta(s)$ with some Euler factors removed) also has a simple pole at $s=1$. The rest of the factors on the RHS are holomorphic at $s=1$. Counting zeros/poles, we see that $L(\chi,1)\neq0$ if $\chi$ is nontrivial.
An analagous result holds for arbitrary abelian extensions of $\Q$. We’ll see a simple case of this.
First note that unique factorization of ideals gives a product formula
where $\mfp$ ranges over all (nonzero) prime ideals of $\ints K$. Now, fix a quadratic number field $K=\Q(\sqrt d)$ ($d\not\in\bits$ squarefree). Recall the splitting behavior of primes in $K$. Let ($k\in\Z$, and $p$ an odd prime)
be the legendre symbol, and let $D=\disc(K/\Q)$ (i.e. $d$ if $d\equiv1\pmod4$ and $4d$ otherwise). Then, for an odd prime $p$, we have
Because of this, we choose to extend the legendre symbol via
We want to relate this splitting behavior to the Euler product for $\zeta_K(s)$. Consider an odd prime $p\in\Z$. If $p$ is inert, this it will contribute (since $p$ prime with norm $p^2$)
to $\zeta_K(s)$. If $p$ splits, then it will contribute (since there are two primes above $p$, each with norm $p$)
to $\zeta_K(s)$. Finally, if $p$ is ramified, then it will contribute (since there’s one prime over $p$ with norm $p$)
to $\zeta_K(s)$. Thus, we see that
where the product is taken over rational primes $p\in\Z$. This looks like it is saying $\zeta_K(s)=\zeta(s)L(\legendre D\cdot,s)$, but is $\chi(n)=\legendre Dn$ even a Dirichlet character?
Yes, it is (since $D\equiv0,1\pmod4$). First, for $n=p_1^{e_1}\dots p_g^{e_g}$, we define
so $\chi(n)$ is at least welldefined, if not a Dirichlet character. It is multiplicative by design, so to show that it is a Dirichlet character (mod $D$), we only need to know that $\chi(n)$ only depends on the value of $n\pmod D$. I must admit that I do not know how to show this ^{16}, but taking its truth for granted for a bit, this gives $\zeta_K(s)=\zeta(s)L(\chi,s)$. If one takes $D=(1)^{(q1)/2}q$ for $q$ an odd prime, then there is only one quadratic Dirichlet character mod $D$ (i.e. one nontrivial homomorphism $U(D)\to{\pm1}$) since $\units{(\zmod q)}$ is cyclic. This character is $\chi(n)=\legendre nq$, but we’ve just shown that $\chi(n)=\legendre Dn$ is also a character. Thus,
which is the law of quadratic reciprocity. ^{17}

The first time I saw this theorem, I thought it was the kind of dry, technical result that almost never shows up in the wild; I was wrong. ↩

You may object that expanding out the RHS let’s you pick a term with infinitely many prime factors, but this is a nonissue because those’ll all multiply out at 0, so we good. ↩

Actually, I have a guess. The theta function has terms that look like $e^{s}$ while the zeta function has terms that look like $(1/n)^s$. If you want to exploit the theta function’s functional equation, then you need some kind of bridge between the two; this leads you to the Gamma function which has both a $e^{\mrm{blah}}$ factor and a $\mrm{blah}^s$ factor. In fancier words, to introduce terms that help you relate the theta function to the zeta function you do something like take the Mellin transform of $s\mapsto e^{\pi n^2s}$ (I say something like because I think we technically end up using the Mellin transform of $s\mapsto e^{\pi n^2s^2}$ instead (and then substitute $u=x^2$), but potato potato) ↩

I think I’ve also seen this called the completed zeta function, but don’t quote me on that ↩

It’s a little known that this is something Dirichlet investigated to get his spirits up after failing to achieve his one, true goal: wiping out half of all life in the universe ↩

Really less a “dive” than a “dip our toes in the water” ↩

actually, holomorphic when the character is nontrivial ↩

See the end of my “Fourier Analysis” post if you don’t know what this is ↩

In hindsight, it would have been better to define $f(x) = (Nx)^{\eps}e^{\pi(Nx)^2s/N}$, and then apply poisson summation to get $\sum_{k\in\Z}f(k+a) = \sum_{k\in\Z}F(f)(k)e^{2\pi ina}$
PS: –I really wish latex worked in these footnotes– (turns out dreams do come true) ↩ 
If you see a mistake somewhere, let me know ↩

At the very least, letting $N=1$ and $\chi=1$ (so $\eps=0$) recovers the functional equation for the classic theta function, as it should ↩

completed L function? ↩

Proving this is not too hard, but would take too much time. You can probably find a proof in like Davenport or something. ↩

But don’t worry, we will prove something involving primes. ↩

This is not the standard way to do so ↩

More accurately any proof I’ve seen of this uses quadratic reciprocity, and I don’t want to use that for reasons that will become apparent soon. ↩

I must admit, this whole section turned out more dubious than I intended. I wanted to present a (clean) proof of qudratic reciprocity, but it’s unclear to me how much machinary/grunt work one would need to turn this outline into a rigorous, noncircular proof ↩